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1.2 Graphs of Equations Ex. 1 Sketch the graph of the line y = -3x + 5 x y -2 0 1 2 Complete the t-graph.

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Presentation on theme: "1.2 Graphs of Equations Ex. 1 Sketch the graph of the line y = -3x + 5 x y -2 0 1 2 Complete the t-graph."— Presentation transcript:

1 1.2 Graphs of Equations Ex. 1 Sketch the graph of the line y = -3x + 5 x y -2 0 1 2 Complete the t-graph

2 x and y-intercepts To find the x-intercepts, let y be 0 and solve the equation for x. To find the y-intercept, let x be 0 and solve the equation for y. Ex. 2 Find the x- and y-intercepts of the graph of x = y 2 - 3 x-int. Let y = 0 x = 0 2 – 3 x = -3 y-int. Let x = 0 0 = y 2 - 3 3 = y 2 (-3,0)

3 Ex. 3Find the x- and y-intercepts of y = x 3 – 4x x-int. Let y = 0 0 = x 3 – 4x Factor 0 = x(x – 2)(x + 2) 0 2 -2 (0,0) (2,0) (-2,0) y-int. Let x = 0 y = 0 3 – 4(0) = 0 (0,0)

4 Symmetry y-axis x-axisorigin (x,y)(-x,y)(x,y) (x,-y) (x,y) (-x,-y) (-x,y), (x,-y), and (-x,-y) are the key points in determining symmetry.

5 Check for symmetry with respect to both axes and the origin. Ex. 4 xy 3 + 10 = 0 Plug in the three ordered pairs. If you can get it to look like the original equation, it has that symmetry. y-axis (-x,y) (-x)y 3 + 10 = 0 -xy 3 + 10 = 0 Is this, or can we get this to look like the original? No. x-axis (x,-y)x(-y) 3 + 10 = 0 -xy 3 + 10 = 0Not like the original. origin (-x,-y) (-x)(-y) 3 + 10 = 0 xy 3 + 10 = 0 This graph has origin symmetry. Day 1

6 Standard Form of the Equation of a Circle (x – h) 2 + (y – k) 2 = r 2 r Center (h,k)Point on the circle (x,y)

7 Ex. 5The point (3,4) lies on a circle whose center is at (-1,2). Find an equation of the circle. First, find the radius of the circle. Write the equation. (x + 1) 2 + (y – 2) 2 = 20

8 Ex. 6Completing the Square to find a circle’s center and radius. 4x 2 + 4y 2 + 20x – 16y + 37 = 0First, divide by 4 x 2 + 5x + y 2 – 4y = Now complete the square. x 2 + 5x + y 2 – 4y =+ 4


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